Lecture 11 (Application of Integration) Areas between Curves Let and be continuous and on. Let s look at the region between and on.

Transcription

1 Lecture 11 (Application of Integration) Areas between Curves Let and be continuous and on. Let s look at the region between and on.

2 Definition: The area of the region bounded by the curves and, and the lines, where and are continuous and on, is defined as Example: Find the area of the region bounded by and.

3 Example: Find the area of the region enclosed by curves and.

4 Example: Find the area of the region enclosed by,,, and.

5 In general, the area between the curves and and between and is given by Why?

6 Example: Find the area of the region bounded by and.

7 Very often the region is bounded by functions of y instead of x, i.e. and. Then if on, we define the area as

8 Example: Sketch the region enclosed by and, and find its area.

9 Example: Find the area of the triangle with the given vertices: (0, 0), (3, 1), (1, 2).

10 Example: Find the area enclosed by and

11 Example: Find the area enclosed by and from to.

12 Volumes To find volumes, we shall use similar procedure as with areas. We shall start with a simple example: a cylinder. It is known that. To prove it, we cut the cylinder into thin disks (cross-sections). The cross-sectional area is denoted by the volume of each disk is given by.. Thus, So. Then

13 We can use similar principle to find volumes of many other solids: Definition: Let S be a solid that lies between and. If the crosssectional area of S (perpendicular to the x-axis) is, then the volume of S is

14 Example: Find the volume of a sphere of radius r.

15 Example: Find the volume of a pyramid with a squared base of side a and height h.

16 Volumes of Solids of Revolutions Volumes of revolution are obtained by revolving a two-dimensional region about an axis of rotation. Consider the graph of a function defined on. Suppose that the region between and the x-axis is revolved about the x-axis 1 : 1

17 Example: Find the volume of the solid of revolution formed by rotating the region bounded by the x-axis and the graph of from to, about the x-axis.

18 Example: Find the volume of the solid of revolution formed by rotating the region bounded by the x-axis and the graph of, from to, about the x-axis.

19 Example 2 : Find the volume of a solid of revolution generated by between and revolving around the y-axis. 2

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21 Example: Find the volume of the solid obtained by rotating about the x-axis the region bounded by the curves and. The cross-section of this type is called a washer. The area of the washer is equal to the area of the outer disk minus the area of the inner disk:

22 Volume by washers: If the region below, above, and between the lines and is revolved around the x-axis, then the volume of the resulting solid is given by

23 Example 3 : Determine the volume of the solid obtained by rotating the portion of the region bounded by and that lies in the first quadrant about the y-axis. 3

24 Example: Find the volume of the solid obtained by rotating the region enclosed by and about the line.

25 Volumes by Cylindrical Shells Unfortunately, we can't always use discs or washers. Example 4 : Determine the volume of the solid obtained by rotating the region bounded by and the x-axis about the y-axis. Let's try to use washers. 4

26 Several problems arise: Both the inner and outer radius are defined by the same function. In order to use washers we would need to put this function into the form. That is NOT easy in general for a cubic polynomial and in other cases may not even be possible to do. The last problem with rings in this case is not so much a problem as it's just added work. If we were to use rings the limit would be y limits and this means that we will need to know how high the graph goes. To this point the limits of integration have always been intersection points that were fairly easy to find. However, in this case the highest point is not an intersection point, but instead a maximum. Finding it will take some time.

27 So we need to find a different way to cut the solid that will give us a crosssectional area that we can work with. One way of doing this is to think of our solid as a lump of cookie dough and instead of cutting it perpendicular to the axis of rotation we center a cylindrical cookie cutter on the axis of rotation and push it down into the solid. Like this: This will give us a cylindrical shell and we can easily find its surface area:

28 The first cylinder will cut into the solid at and as we increase x to we will completely cover both sides of the solid since expanding the cylinder in one direction will automatically expand it in the other direction as well. The volume of this solid is then, The method used here is called the method of cylindrical shells.

29 Example: Determine the volume of the solid obtained by rotating the region bounded by and about the line.

30 In this example both methods can be used (shells or washers)

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